Make button open link – Swift

What your code shows is the action that would occur once the button is tapped, rather than the actual button. You need to connect your button to that action.

(I’ve renamed the action because GoogleButton is not a good name for an action)

In code:

override func  viewDidLoad() {
  super.viewDidLoad()

  googleButton.addTarget(self, action: "didTapGoogle", forControlEvents: .TouchUpInside)
}

@IBAction func didTapGoogle(sender: AnyObject) {
  UIApplication.sharedApplication().openURL(NSURL(string: "http://www.google.com")!)
}

In IB:

Edit: in Swift 3, the code for opening a link in safari has changed. Use UIApplication.shared().openURL(URL(string: "http://www.stackoverflow.com")!) instead.

Edit: in Swift 4
UIApplication.shared.openURL(URL(string: "http://www.stackoverflow.com")!)